二叉搜索树转换为双向链表

要求空间复杂度为O(1)，两种方法：

1、不严格的O(1)，用了递归

2、严格的递归，用了Morris中序遍历（中序遍历在tempNode->right == root时，不需要清零，因为此时prevNode正好等于tempNode）

struct TreeNode
{
    int val;
    TreeNode *left, *right;
    TreeNode(int _val) : val(_val), left(nullptr), right(nullptr)
    {

    }
};

void doubleLinkList(TreeNode *root, TreeNode *&prevNode)
{
    if (root == nullptr)
        return;

    doubleLinkList(root->left, prevNode);

    prevNode->right = root;
    root->left = prevNode;
    prevNode = root;

    doubleLinkList(root->right, prevNode);
}

void doubleLinkList2(TreeNode *root, TreeNode *prevNode)
{
    while (root)
    {
        if (root->left == nullptr)
        {
            prevNode->right = root;
            root->left = prevNode;
            prevNode = root;
            root = root->right;
        }
        else
        {
            TreeNode *tempNode = root->left;
            while (tempNode->right != nullptr && tempNode->right != root)
                tempNode = tempNode->right;

            if (tempNode->right == nullptr)
            {
                tempNode->right = root;
                root = root->left;
            }
            else
            {
                root->left = prevNode;
                prevNode = root;
                root = root->right;
            }
        }
    }
}


int main()
{
    TreeNode *root = new TreeNode(5);
    root->left = new TreeNode(3);
    root->left->left = new TreeNode(1);
    root->right = new TreeNode(6);

    TreeNode dummy(0);
    TreeNode *prevNode = &dummy;
    //doubleLinkList(root, prevNode);
    doubleLinkList2(root, prevNode);
    
    if (dummy.right != nullptr)
        dummy.right->left = nullptr;

    return 0;
}